The probability distribution of random variable X is given by: X 1 2 3 4 5 P ( X ) K 2K 2K 3K K Let p = P ( 1

Correct answer: 0 ΣP ( X ) = 1 ⇒ K + 2K + 2K + 3K + K = 1 K = (1)/(9) P = P ( 1 P = ((2K)/(9K))/((K)/(9K) + (2k)/(9K)) = (2)/(3) 5P = 5 × (2)/(3) = λ × (1)/(9) ⇒ λ = 30

JEE Main 202127 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2021 · 27 August · Evening Shift · Mathematics · Probability

The probability distribution of random variable X is given by: X 1...

The probability distribution of random variable $X$ is given by:

$X$	$1$	$2$	$3$	$4$	$5$
$P ( X )$	$K$	$2K$	$2K$	$3K$	$K$

Let $p = P ( 1 < X < 4 ∣ X < 3 )$ . If $5p = \lambdaK$ , then $\lambda$ is equal to

Correct answer

Step-by-step solution

ΣP ( X ) = 1

⇒ K + 2K + 2K + 3K + K = 1

K = \dfrac{1}{9}

P = P ( 1 < X < 4 ∣ X < 3 ) = \dfrac{P ( X = 2 )}{P ( X < 3 )}

P = \dfrac{\dfrac{2K}{9K}}{\dfrac{K}{9K} + \dfrac{2k}{9K}} = \dfrac{2}{3}

5P = \lambdaK

5 \times \dfrac{2}{3} = \lambda \times \dfrac{1}{9} ⇒ \lambda = 30

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